# Why can't I add the period to find the next two solutions?

#### davedave

Consider tan(2x) = $$\displaystyle \sqrt3$$ for 0 $$\displaystyle \leq x < 2\pi$$

reference angle x = $$\displaystyle \tan^{-1}$$($$\displaystyle \sqrt{3}$$) = $$\displaystyle \frac{\pi}{3}$$

In the given domain 0 $$\displaystyle \leq x < 2\pi$$, the positive ratio of tangent is in quadrants I and III. So the two standard position angles in this interval are

2x = $$\displaystyle \frac{\pi}{3}$$ and $$\displaystyle \frac{4\pi}{3}$$

Now, divide both sides by 2. Then I get

x = $$\displaystyle \frac{\pi}{6}$$ and $$\displaystyle \frac{2\pi}{3}$$ --------- (*)

The period of tangent is: period = $$\displaystyle \frac{\pi}{b}$$

where b = 2 in tan(2x).

So the period is $$\displaystyle \frac{\pi}{2}$$.

Then, add this period to each solution of x in the first cycle to get the next two solutions in the second cycle. So I get

x = $$\displaystyle \frac{2\pi}{3}$$ and $$\displaystyle \frac{7\pi}{6}$$

Therefore, the solutions are x=$$\displaystyle \frac{\pi}{6}$$, $$\displaystyle \frac{2\pi}{3}$$, $$\displaystyle \frac{2\pi}{3}$$ and $$\displaystyle \frac{7\pi}{6}$$

The two middle solutions are the same.

The answer key answer says x = $$\displaystyle \frac{\pi}{6}$$, $$\displaystyle \frac{2\pi}{3}$$, $$\displaystyle \frac{7\pi}{6}$$ and $$\displaystyle \frac{5\pi}{3}$$.

I actually found my mistake. If I added $$\displaystyle \pi$$ instead of $$\displaystyle \frac{\pi}{2}$$ in (*) above, I would get the right solutions in the second cycle. But, I don't understand why you have to do that, since the period of tan(2X) is $$\displaystyle \frac{\pi}{b}$$ = $$\displaystyle \frac{\pi}{2}$$.

I am very confused. Can someone explain this? Thanks a lot.

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#### skeeter

Math Team
$\tan(2x) = \sqrt{3}$

$0 \le x < 2\pi \implies 0 \le 2x < 4\pi$

$2x = \dfrac{\pi}{3} \, , \, \dfrac{4\pi}{3} \, , \, \dfrac{7\pi}{3} \, , \, \dfrac{10\pi}{3}$

$x = \dfrac{\pi}{6} \, , \, \dfrac{2\pi}{3} \, , \, \dfrac{7\pi}{6} \, , \, \dfrac{5\pi}{3}$

note ...

$\dfrac{\pi}{6} + \dfrac{\pi}{2} = \dfrac{2\pi}{3}$

$\dfrac{\pi}{6} + 2 \cdot \dfrac{\pi}{2} = \dfrac{7\pi}{6}$

$\dfrac{\pi}{6} + 3 \cdot \dfrac{\pi}{2} = \dfrac{5\pi}{3}$

... adding the modified period does work

#### greg1313

Forum Staff
Note that the difference between $\dfrac{\pi}{6}$ and $\dfrac{2\pi}{3}$ is $\dfrac{\pi}{2}$.

If you had added $\dfrac{\pi}{2}$ to $\dfrac{2\pi}{3}$ twice you would have come up with the two remaining solutions, hence the period of $\dfrac{\pi}{2}$.

$$\tan(2x)=\sqrt{3}$$
$$2x=\dfrac{\pi}{3}+k\pi$$
$$x=\dfrac{\pi}{6}+k\dfrac{\pi}{2},\quad k\in\mathbb{Z}$$

#### skipjack

Forum Staff
Then, add this period to each solution of x in the first cycle to get the next two solutions in the second cycle.
You cover two periods in each cycle, so adding the period to your first solution, $$\displaystyle \frac{\pi}{6}$$, gives your second solution in the same cycle, $$\displaystyle \frac{2\pi}{3}$$. Adding the period to your second solution gives the first solution in the second cycle, $$\displaystyle \frac{7\pi}{6}$$. Adding the period once more will then give the second solution in the second cycle, $$\displaystyle \frac{5\pi}{3}$$.

As the cycle length is a multiple of the period for this question, you can alternatively add the cycle length, $\pi$, to each of your initial solutions to get the solutions in the second cycle.

To avoid confusion, I recommend that you start by finding the solution(s) in the first period (rather than cycle) and then add the period length repeatedly until the specified domain is exhausted.